# Why aren't coordinates induced vector fields always Killing fields?

We have that $$L_K g_{\mu\nu}=\nabla_\mu K_\nu + \nabla_\nu K_\mu$$

A vector field $$K$$ is a Killing field if $$L_K g_{\mu\nu}=0$$, but consider the coordinate induced vector field $$\partial_\alpha$$, we have

$$(\partial_\alpha)_\nu=g_{\lambda\nu}(\partial_\alpha)^\lambda= g_{\lambda\nu}\delta^\lambda_{\,\,\alpha}=g_{\alpha\nu}$$

Thus by the compatibility condition of the Levi Civita connection, i.e. $$g_{ij;k}=0$$ for all $$i,j,k$$ $$L_{\partial_\alpha}g_{\mu\nu}=0$$

for all $$\alpha$$.

This is of course nonsensical, because it would imply that velocities are always conserved in every direction, hence there's never acceleration... where is the huge fault in my reasoning?

EDIT: it gets worse: any non vanishing vector field on a smooth manifold can be expressed as $$\partial_1$$ in a suitable chart

• What is $(\partial_\alpha)_\nu$ suppose to mean? Of course the statement is wrong. Take any example you wish, where the components of the metric depend on a given coordinate, then the tangent field corresponding to that coordinate is not Killing. – MBN Dec 3 '18 at 13:11
• I suggest taking $K=\partial_r$ and $g_{\mu\nu}$ the Schwarzschild metric, and seeing where and how your reasoning fails. – mmeent Dec 3 '18 at 13:45
• @MBN $(\partial_\alpha)^\nu$ is the component $\nu$ of the vector field $\frac{\partial}{\partial x^\alpha}$, i.e. $\mathrm{d}x^\nu \partial_\alpha=\delta^\nu_{\,\alpha}$ – user2723984 Dec 3 '18 at 14:52
• @mment do you mean calculating the Lie derivative directly without using the covariant derivative? – user2723984 Dec 3 '18 at 15:01
• @mment Is my computation that shows $K_\mu=g_{\mu\alpha}$ incorrect or did I misunderstand the Levi Civita condition? – user2723984 Dec 3 '18 at 15:24

A vector field K is a Killing field if the Lie derivative with respect to K of the metric g vanishes. In your demonstration you assume as vector the partial derivative and then in the R.H.S. of the equation you show up with a tensor, i.e. the metric. It is inconsistent. What your demonstration defines are the covariant components of the partial derivative as vector, not the components of the metric tensor. So, the compatibility condition is not applicable.

Note:
Given a metric $$g_{\mu \nu}$$, a Killing field $$K = \partial_\lambda$$ exists if all of the components of the metric are independent of the coordinate $$x^\lambda$$. However there may be hidden symmetries not so manifest.

Further:
The metric compatibility is not applicable because the covariant derivative of a vector (one index) differs from the covariant derivative of a two index tensor.
$$\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu \sigma} V^\sigma$$ Eq. (1) covariant derivative of a vector
$$\nabla_\mu T^{\nu \lambda} = \partial_\mu T^{\nu \lambda} + \Gamma^\nu_{\mu \sigma} T^{\sigma \lambda} + \Gamma^\lambda_{\mu \sigma} T^{\nu \sigma}$$ Eq. (2) covariant derivative of a two index tensor
The structure of the covariant derivative is different. Even if the components of the partial derivative vector are formally the same as the components of the metric tensor, they are worked out according to Eq. (1) and not to Eq. (2), which would assure the compatibility.
(Just note that I assumed both vector and tensor as described in contravariant components. If you have in covariant components there is a $$-$$ sign in front of the $$\Gamma's$$ and the indices change up/down).

• Thank you for the answer. I'm not sure I understand. I'm new to GR and I thought that if $K$ is a vector $g_{\mu\nu}K^\nu$ was the definition of $K_\mu$. If I understood correctly, to obtain the covariant form of $(\partial_\alpha)^\nu$ I can't simply contract with the metric. But then what is the covariant form and how do I obtain it? How do I see that for the Schwarzschild metric, $\partial_\phi$ and $\partial_t$ are Killing fields but $\partial_r$ isn't? – user2723984 Dec 3 '18 at 17:57
• I restated my answer. I added also a note on a manifest way to understand whether the partial derivative is a Killing field. – Michele Grosso Dec 4 '18 at 16:59
• "What your demonstration defines are the covariant components of the partial derivative as vector, not the components of the metric tensor" In my "demonstration" the covariant components of the partial derivative as a vector turn out to be 4 components of the metric, they're the same mathematical object, I don't see why the compatibility condition doesn't apply. I think I need to revise some differential geometry. I noticed the more explicit condition by using the direct expression of the Lie derivative, it's still not clear to me what exactly goes wrong using the covariant derivatives – user2723984 Dec 5 '18 at 7:37
• I added the section "Further" in my answer to explain why. – Michele Grosso Dec 5 '18 at 17:24